We study this model in the particle-hole symmetric case, corresponding to \(\epsilon _{\tau } = - \frac{U_\tau }{2}\)  and
symmetric conduction bands relative to the \(\omega =0\)  Fermi level) in order to have exactly one hole in each impurity orbital. In the Kondo regime, \(-\epsilon _{\tau } \gg V^2_\tau /W\) , these
two holes are ferromagnetically coupled by the first Hund's rule, \(J_H > 0\) , yielding an effective \(S=1\)  at the impurity site.
Due to symmetry considerations of the involved 3d orbitals, we take the same Hubbard repulsion \(U\)  for each impurity orbital.
This implies that both single-orbital energies \(\epsilon _\tau \)  should be degenerate in the particle-hole symmetric case, while ab initio calculations [1]} give
an energy difference of the order of 1 eV. In the Kondo regime this difference is irrelevant.
